To solve the equation \( \frac{k^5 \cdot k^2}{k^{6x}} = (k^2)^{-2} \), we first simplify both sides.
Starting with the left-hand side:
\[ \frac{k^5 \cdot k^2}{k^{6x}} = \frac{k^{5 + 2}}{k^{6x}} = \frac{k^7}{k^{6x}} = k^{7 - 6x} \]
Now, for the right-hand side:
\[ (k^2)^{-2} = k^{2 \cdot (-2)} = k^{-4} \]
Now the equation is:
\[ k^{7 - 6x} = k^{-4} \]
Since the bases are the same, we can set the exponents equal to each other:
\[ 7 - 6x = -4 \]
Now, we solve for \( x \):
- Subtract 7 from both sides:
\[ -6x = -4 - 7 \] \[ -6x = -11 \]
- Divide by -6:
\[ x = \frac{11}{6} \]
Therefore, the missing exponent \( x \) that would make the equation true is
\[ \boxed{\frac{11}{6}} \]